I have trouble understanding what is a germ in a topological space.
I have to mention that i don't have knowledge about this concept in any other area (complex analysis or others), so please don't answer by this means.
The definitions that have been given to me are this:
Let $(X,\tau_1)$ and $(Y,\tau_2)$ be topological spaces, and $x\in X$.
Let $\xi(x)$ denote the set of all neighbourhoods of $x$.
- Let $U,V\in\xi(x)$, $f:U\rightarrow Y$ and $g:V\rightarrow Y$ two maps.
We say that $f$ and $g$ are germ-equivalent at $x$ if $\exists W\in\xi(x)\;:\;W\subseteq U\cap V,\;\forall y\in W,\;f(y)=g(y)$. - Let $A,B\subseteq X$.
We say $A$ and $B$ are germ-equivalent at $x$ if $\exists U\in\xi(x):A\cap U=B\cap U$.
This two relations define two equivalence relations.
And what i wanted to prove is that $[A]_x=[B]_x$ $\Leftrightarrow$ $[\chi_A]_x=[\chi_B]_x$.
If this is not true i will be grateful if someone can provide a counterexample.
I have to mention that i'm newbie in this so maybe i have made some mistakes.
The statement is true. Here's a proof.
"$\Longrightarrow$"
Suppose $A$ and $B$ are germ-equivalent at $x$, so $A\cap U = B\cap U$ for some $U \in \xi (x)$. Then clearly $\mathbb{1}_A(y) = \mathbb{1}_B(y)$ for all $y \in U$. By definition we say that $\mathbb{1}_A$ is germ-equivalent to $\mathbb{1}_B$ at $x$.
"$\Longleftarrow$"
Now suppose that $\mathbb{1}_A(y) = \mathbb{1}_B(y)$ for all $y \in U$ for some $U \in \xi(x)$. Then clearly $A \cap U = B\cap U$, and by definition, $A$ is germ-equivalent to $B$ at $x$.
To put the argument more succinctly: $$A \cap U = B \cap U \Longleftrightarrow \mathbb{1}_A(y) = \mathbb{1}_B(y), \,\, \forall y \in U$$
From which it follows that one germ-equivalence holds if and only if the other does.
Let me know if you need me to provide more details in either direction of the proof to be convincing.